Question

T (hours) 2 5 9 11 12

L(t) (car per hours) 15 40 24 68 18


The rate at which cars enter a parking lot is modeled by E(t)=30+5(t-2)(t-5)e⁻⁰.²ᵗ. The rate at which cars leave the parking lot is modeled by the differentiable function L. Selected values of L (£) are given in the table above. Both E(t) and L(t) are measured in cars per hour, and time t is measured in hours after 5A.M. (t=0). Both functions are defined for 0≤ t≤ 12.

(a) What is the rate of change of E(t) at time t=7 ? Indicate units of measure.

Answer 1

a. The rate of change at t = 2 is approximately 8 cars per hour.

b. The total number of cars entering the parking lot from t = 2 to
t = 12 is 155 .

c. Using a trapezoidal sum, the approximation of the total accumulation of cars during the specified time intervals is 155.

d. The total revenue collected from cars entering the parking lot from
t = 2 to t = 12 is $919 .

a. To estimate the rate of change, we use the average rate of change between adjacent time points:

`\[ L'(2) \approx (40 - 15)/(5 - 2) = 8 \, \text{cars per hour} \]`

b. The number of cars entering the parking lot from t = 2 to t = 12 is the sum of the areas of the trapezoids formed by adjacent points:

`\[ \text{Cars} = (15 + 2 \cdot 40 + 24 + 2 \cdot 68 + 18)/(2) = 155 \]`

c. Using a trapezoidal sum, the approximation of the total accumulation of cars during the specified time intervals is 155.

d. To find the total revenue, we multiply the number of cars by the corresponding rate and sum these values:

`\[ \text{Revenue} = 155 \cdot 5 + 18 \cdot 8 = 775 + 144 = 919 \]`

Therefore, the answer is $919.